2
$\begingroup$

I know that for quaternions, $$i^2=j^2=k^2=ijk=-1$$ I've tried to understand this intuitively as having $i$, $j$ and $k$ represent a rotation about each of three axes. But when I do a bit of manipulation with the quaternions, I ran into a problem. If $$i^2=ijk$$ shouldn't $$i=jk \text{ ?}$$ But because $$i^2=-1$$ $$(jk)^2=-1$$ and $$j^2k^2=-1$$ which is obviously not true: $j^2k^2$ should equal $1$.

Where is my error?

By the way, Wikipedia notes that quaternionic multiplication is non-commutative. Does that have something to do with it?

$\endgroup$
  • 3
    $\begingroup$ as mentioned below, your error is that $(jk)^2 \neq j^2k^2$ but instead $(jk)^2 = (jk)(jk) = jkjk$ $\endgroup$ – JMoravitz Nov 12 '14 at 2:23
3
$\begingroup$

I view quaternions as reading. The order is $\{1, i, j, k\}$ so we read left to right. That is, $ij = k$ since that is left to right whereas $ji = -k$ since this is right to left so it is reversed, $-1$. $$ (jk)^2 = jkjk = ii = -1 $$ Also note that $$ j^2k^2 = jjkk = jik = -kk = 1\neq jkjk = ii = -1 $$ Does this help?

$\endgroup$
  • 2
    $\begingroup$ So it hinges on $(jk)^2 \neq j^2k^2$? $\endgroup$ – HDE 226868 Nov 12 '14 at 2:22
  • 2
    $\begingroup$ @HDE226868 you changed the order. $jjkk = jik = -kk = 1$ is not $jkjk$ $\endgroup$ – dustin Nov 12 '14 at 2:23
1
$\begingroup$

There is another way of looking at quaternions - they are identical to matrices of a certain kind. Quaternions of the form $bi + cj + dk + a$ can be represented by matrices of the form $$ \begin{bmatrix} a+bi & c+di \\ -c+di & a-bi \end{bmatrix} $$ where the $i$ inside the matrix is $\sqrt{-1}$. The matrix representing the quaternion $i$ is $$ \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} $$ Write down $jk$ in this form $$ jk= \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} $$ which is clearly $i$, not $-i$.

The rule is do nothing with quaternions that you cannot do with matrices of the form given above and you'll be ok.

Edited to add: If you think of $(jk)^2$ in terms of matrix multiplication, $$(jk)^2=jkjk=ii= \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} =-1 $$ But $j^2k^2$ = $jjkk = jik = $

$$ \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} =1 $$

$\endgroup$
  • $\begingroup$ I'm guessing that this goes along with the non-commutability? (i.e. for matrices, it is sometimes the case that $AB \neq BA$) $\endgroup$ – HDE 226868 Nov 12 '14 at 2:39
  • $\begingroup$ @HDE226868 jk is indeed equal to i, for the reasons given above. $\endgroup$ – user_of_math Nov 12 '14 at 2:44
  • $\begingroup$ @HDE226868 read the quaternion tag. It will tell you it is noncommutative division algebra. $\endgroup$ – dustin Nov 12 '14 at 2:59
  • $\begingroup$ @HDE226868 Yeah, quaternion algebra is associative but not commutative. $\endgroup$ – user_of_math Nov 12 '14 at 3:01
  • $\begingroup$ @dustin Yes, I knew that, but wasn't sure how it applied here. $\endgroup$ – HDE 226868 Nov 12 '14 at 3:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.